Help with derivative of y= ln x

So I have this problem solved algebraically, but need to use derivatives...Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)= -ln(1/a * 1/b)= -[ln(1/a) + ln(1/b)]= -ln(1/a) + -ln(1/b) = ln(a) + ln(b) this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...

So I have this problem solved algebraically, but need to use derivatives...Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)= -ln(1/a * 1/b)= -[ln(1/a) + ln(1/b)]= -ln(1/a) + -ln(1/b) = ln(a) + ln(b) this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...

So I have this problem solved algebraically, but need to use derivatives...Explain how to use the derivative formula to prove that ln(a) + ln(b)= ln(ab).The derivative formula can be used to prove ln(a) + ln(b) = ln(ab) in the following way:

ln(ab) = -ln(1/ab)= -ln(1/a * 1/b)= -[ln(1/a) + ln(1/b)]= -ln(1/a) + -ln(1/b) = ln(a) + ln(b) this is what I have come up with, but it is not right...
I need to differentiate both sides with respect to a, treating b as a constant. Because the expressions are equal, they can differ at most by a constant C. I need to show that c=0, while evaluating the function with b=1.
Any help out there would be greatly appreciated. Thanks for looking...